MATHEMATICS

\displaystyle \text{Series SGN/1} \hspace{1.0cm} | \hspace{1.0cm} \text{Q. P. Code 65/1/3} \hspace{1.0cm} | \hspace{1.0cm} \text{Set 3 }    

\displaystyle \text{Time Allowed : 3 hours} \hspace{5.0cm} \text{Maximum Marks : 100 }  

 

\displaystyle \textbf{General Instructions:}
\displaystyle \text{(i) All questions are compulsory.}
\displaystyle \text{(ii) The question paper consists of 29 questions divided into four sections A, B, C }
\displaystyle \text{and D. Section A comprises of 4 questions of one mark each, Section B comprises of 8 }
\displaystyle \text{questions of two marks each, Section C comprises of 11 questions of four marks each }
\displaystyle \text{and Section D comprises of 6 questions of six }   \text{marks each.}
\displaystyle \text{(iii) All questions in Section A are to be answered in one word, one sentence or as per }
\displaystyle \text{the exact requirement of the question.}
\displaystyle \text{(iv) There is no overall choice. However, internal choice has been provided in 3 questions of }
\displaystyle \text{four marks each and 3 questions }   \text{of six marks each. You have to attempt only one of the}
\displaystyle \text{alternatives in all such questions.}
\displaystyle \text{(v) Use of calculators is not permitted. You may ask for} \text{logarithmic tables, if required.}

\displaystyle \textbf{SECTION - A}
\displaystyle \text{Question numbers 1 to 4 carry 1 mark each.}

\displaystyle \textbf{1. } \text{If a * b denotes the larger of 'a' and 'b' and if a o b = (a * b) + 3, then }
\displaystyle \text{write the value of (5) o (10),}  \text{where * and o are binary operations.}
\displaystyle \text{Answer:}
\displaystyle  \text{(5) o (10)}=(5 \ast 10)+3=10+3=13\ \ (\because\ 10>5)

\displaystyle \textbf{2. } \text{Find the magnitude of each of the two vectors } \overrightarrow{a} \text{ and } \overrightarrow{b}\text{, having the same magnitude }
\displaystyle \text{such that the angle between }   \text{them is } 60^\circ \text{ and their scalar product is } \frac{9}{2}.
\displaystyle \text{Answer:}
\displaystyle \text{As }\overrightarrow{a}\cdot\overrightarrow{b}=|\overrightarrow{a}||\overrightarrow{b}|\cos\theta,\ \text{where }\theta\text{ is angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}
\displaystyle \therefore \frac{9}{2}=|\overrightarrow{a}|^2\cos60^\circ\ \ (\because\ |\overrightarrow{a}|=|\overrightarrow{b}|\ \text{given})
\displaystyle \Rightarrow \frac{9}{2}=|\overrightarrow{a}|^2\cdot\frac{1}{2}\Rightarrow 9=|\overrightarrow{a}|^2
\displaystyle \therefore |\overrightarrow{a}|=3
\displaystyle \text{Hence magnitude of each of the vectors }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ is }3.

\displaystyle \textbf{3. } \text{If the matrix } A = \begin{bmatrix} 0 & a & -3 \\ 2 & 0 & -1 \\ b & 1 & 0 \end{bmatrix} \text{ is skew symmetric, find the }   \text{values of 'a' and 'b'.}
\displaystyle \text{Answer:}
\displaystyle   A=\begin{bmatrix}0 & a & -3\\ 2 & 0 & -1\\ b & 1 & 0\end{bmatrix}\ \text{is skew symmetric.}
\displaystyle \begin{bmatrix}0 & 2 & b\\ a & 0 & 1\\ -3 & -1 & 0\end{bmatrix}=\begin{bmatrix}0 & -a & 3\\ -2 & 0 & 1\\ -b & -1 & 0\end{bmatrix}\ \ (\because\ A^T=-A)
\displaystyle \therefore a=-2,\ b=3

\displaystyle \textbf{4. } \text{Find the value of } \tan^{-1} \sqrt{3} - \cot^{-1}(-\sqrt{3}).
\displaystyle \text{Answer:}
\displaystyle  \tan^{-1}\sqrt{3}-\cot^{-1}(-\sqrt{3})
\displaystyle =\tan^{-1}\sqrt{3}-\left[\pi-\cot^{-1}\sqrt{3}\right]\ \ (\because\ \cot^{-1}(-x)=\pi-\cot^{-1}x)
\displaystyle =\left(\tan^{-1}\sqrt{3}+\cot^{-1}\sqrt{3}\right)-\pi
\displaystyle =\left(\frac{\pi}{3}+\frac{\pi}{6}\right)-\pi=\frac{\pi}{2}-\pi=-\frac{\pi}{2}

\displaystyle \textbf{SECTION - B}
\displaystyle \text{Question numbers 5 to 12 carry 2 mark each.}

\displaystyle \textbf{5. } \text{The total cost C(x) associated with the production of x units of an item is given by }
\displaystyle  C(x) = 0.005x^3 - 0.02x^2 + 30x + 5000.  \text{ Find the marginal cost when 3 units}
\displaystyle \text{are produced, where by}   \text{marginal cost we mean the instantaneous rate of change of}
\displaystyle \text{total cost at any level of output.}
\displaystyle \text{Answer:}
\displaystyle \text{As }C(x)=0.005x^3-0.02x^2+30x+5000
\displaystyle \therefore \text{Marginal cost}=C'(x)
\displaystyle =3(0.005x^2)-0.02(2x)+30=0.015x^2-0.04x+30
\displaystyle \text{Marginal cost when }3\text{ units are produced }=C'(3)
\displaystyle =0.015(3)^2-0.04(3)+30=0.135-0.12+30=30.015

\displaystyle \textbf{6. } \text{Differentiate } \tan^{-1}\left(\frac{1+\cos x}{\sin x}\right) \text{ with respect to } x.
\displaystyle \text{Answer:}
\displaystyle \text{Let }f(x)=\tan^{-1}\left(\frac{1+\cos x}{\sin x}\right)
\displaystyle =\tan^{-1}\left(\frac{2\cos^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}}\right)=\tan^{-1}\left(\frac{\cos\frac{x}{2}}{\sin\frac{x}{2}}\right)
\displaystyle =\tan^{-1}\left(\cot\frac{x}{2}\right)=\tan^{-1}\left[\tan\left(\frac{\pi}{2}-\frac{x}{2}\right)\right]
\displaystyle f(x)=\frac{\pi}{2}-\frac{x}{2}
\displaystyle f'(x)=-\frac{1}{2}

\displaystyle \textbf{7. } \text{Given } A = \begin{bmatrix} 2 & -3 \\ -4 & 7 \end{bmatrix} \text{, compute } A^{-1} \text{ and show that }   \text{2A}^{-1} = 9I - A.
\displaystyle \text{Answer:}
\displaystyle   A=\begin{bmatrix}2 & -3\\ -4 & 7\end{bmatrix}
\displaystyle C_{11}=(-1)^{1+1}(7)=7,\ C_{12}=(-1)^{1+2}(-4)=4,
\displaystyle C_{21}=(-1)^{2+1}(-3)=3,\ C_{22}=(-1)^{2+2}(2)=2
\displaystyle \text{Adj }A=\begin{bmatrix}7 & 4\\ 3 & 2\end{bmatrix}^T=\begin{bmatrix}7 & 3\\ 4 & 2\end{bmatrix}
\displaystyle |A|=2\times 7-(-3)\times(-4)=14-12=2\neq 0
\displaystyle \therefore A^{-1}=\frac{1}{|A|}\text{ adj }A=\frac{1}{2}\begin{bmatrix}7 & 3\\ 4 & 2\end{bmatrix}=\begin{bmatrix}\frac{7}{2} & \frac{3}{2}\\ 2 & 1\end{bmatrix}
\displaystyle \text{L.H.S.}=2A^{-1}=\begin{bmatrix}7 & 3\\ 4 & 2\end{bmatrix}
\displaystyle \text{R.H.S.}=9I-A=9\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}-\begin{bmatrix}2 & -3\\ -4 & 7\end{bmatrix}
\displaystyle =\begin{bmatrix}9 & 0\\ 0 & 9\end{bmatrix}-\begin{bmatrix}2 & -3\\ -4 & 7\end{bmatrix}=\begin{bmatrix}7 & 3\\ 4 & 2\end{bmatrix}
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence verified that }2A^{-1}=9I-A.

\displaystyle \textbf{8. } \text{Prove that: }   \text{3}\sin^{-1} x = \sin^{-1}(3x - 4x^3), \; x \in \left[-\frac{1}{2}, \frac{1}{2}\right].
\displaystyle \text{Answer:}
\displaystyle   3\sin^{-1}x=\sin^{-1}(3x-4x^3),\ x\in\left[-\frac{1}{2},\frac{1}{2}\right]
\displaystyle \text{Let }\sin^{-1}x=\theta\Rightarrow x=\sin\theta
\displaystyle \text{L.H.S.}=3\theta
\displaystyle \text{R.H.S.}=\sin^{-1}(3\sin\theta-4\sin^3\theta)=\sin^{-1}(\sin3\theta)=3\theta
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S.}
\displaystyle \text{Hence, }3\sin^{-1}x=\sin^{-1}(3x-4x^3)

\displaystyle \textbf{9. } \text{A black and a red die are rolled together. Find the conditional probability of }
\displaystyle \text{obtaining the sum 8, given that the red die }   \text{resulted in a number less than 4.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A\text{ be the event of obtaining sum as }8\text{ and }B\text{ be the event that red die resulted}
\displaystyle \text{in a number less than }4.\text{ Let black die be represented by first number in the ordered pair.}
\displaystyle B\text{ has the following outcomes}
\displaystyle (1,1),(2,1),(3,1),(4,1),(5,1),(6,1),(1,2),(2,2),(3,2),(4,2),(5,2),(6,2),
\displaystyle (1,3),(2,3),(3,3),(4,3),(5,3),(6,3)
\displaystyle \text{So, }B\text{ has }18\text{ outcomes}
\displaystyle A\text{ has }(2,6),(3,5),(4,4),(5,3),(6,2)
\displaystyle \text{So, }A\text{ has }5\text{ outcomes}
\displaystyle A\cap B\text{ has }(5,3)\text{ and }(6,2)
\displaystyle \therefore P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{2}{18}=\frac{1}{9}

\displaystyle \textbf{10. } \text{If } \theta \text{ is the angle between two vectors } \widehat{i} - 2\widehat{j} + 3\widehat{k} \text{ and } 3\widehat{i} - 2\widehat{j} + \widehat{k} \text{, find } \sin \theta.
\displaystyle \text{Answer:}
\displaystyle  \overrightarrow{a}\times\overrightarrow{b}=\begin{vmatrix}\widehat{i} & \widehat{j} & \widehat{k}\\ 1 & -2 & 3\\ 3 & -2 & 1\end{vmatrix}
\displaystyle =\widehat{i}((-2)(1)-3(-2))-\widehat{j}(1\cdot 1-3\cdot 3)+\widehat{k}(1(-2)-(-2)3)
\displaystyle =\widehat{i}(-2+6)-\widehat{j}(1-9)+\widehat{k}(-2+6)
\displaystyle =4\widehat{i}+8\widehat{j}+4\widehat{k}
\displaystyle |\overrightarrow{a}\times\overrightarrow{b}|=\sqrt{4^2+8^2+4^2}=\sqrt{16+64+16}=\sqrt{96}=4\sqrt{6}
\displaystyle |\overrightarrow{a}|=\sqrt{1^2+(-2)^2+3^2}=\sqrt{14},\ |\overrightarrow{b}|=\sqrt{3^2+(-2)^2+1^2}=\sqrt{14}
\displaystyle |\overrightarrow{a}\times\overrightarrow{b}|=|\overrightarrow{a}||\overrightarrow{b}|\sin\theta
\displaystyle \Rightarrow \sin\theta=\frac{4\sqrt{6}}{\sqrt{14}\cdot\sqrt{14}}=\frac{4\sqrt{6}}{14}=\frac{2\sqrt{6}}{7}

\displaystyle \textbf{11. } \text{Find the differential equation representing the family of }   \text{curves } y = a e^{bx + 5} \\ \text{ where a and b are arbitrary constants.}
\displaystyle \text{Answer:}
\displaystyle   y=ae^{bx+5}\ \ ...(i)
\displaystyle \Rightarrow y'=bae^{bx+5}=by\ \ ...(ii)
\displaystyle \Rightarrow y''=b^2ae^{bx+5}=b^2y
\displaystyle \Rightarrow y''=\left(\frac{y'}{y}\right)^2 y\ \ \text{[By using (ii)]}
\displaystyle \Rightarrow yy''-(y')^2=0
\displaystyle \text{This represents the differential equation representing the family of curves}
\displaystyle y=ae^{bx+5},\ \text{where }a\text{ and }b\text{ are arbitrary constants.}

\displaystyle \textbf{12. } \text{Evaluate: } \int \frac{\cos 2x + 2\sin^2 x}{\cos^2 x} \, dx
\displaystyle \text{Answer:}
\displaystyle   \int \frac{\cos2x+2\sin^2x}{\cos^2x}dx
\displaystyle =\int \frac{2\cos^2x-1+2(1-\cos^2x)}{\cos^2x}dx \\ \text{[Put }\cos2x=2\cos^2x-1,\ \sin^2x=1-\cos^2x]
\displaystyle =\int \frac{1}{\cos^2x}dx=\int \sec^2x\,dx=\tan x+C

\displaystyle \textbf{SECTION - C}
\displaystyle \text{Question numbers 13 to 22 carry 4 marks each.}

\displaystyle \textbf{13. } \text{If } y = \sin(\sin x)\text{, prove that:  }   \frac{d^2 y}{dx^2} + \tan x \frac{dy}{dx} + y \cos^2 x = 0.
\displaystyle \text{Answer:}
\displaystyle \text{As }y=\sin(\sin x)
\displaystyle \Rightarrow \frac{dy}{dx}=\cos(\sin x)\cos x\ \ ...(i)
\displaystyle \frac{d^2y}{dx^2}=\cos(\sin x)(-\sin x)-\cos^2x\sin(\sin x)
\displaystyle \text{Now }\frac{d^2y}{dx^2}+\tan x\frac{dy}{dx}+y\cos^2x
\displaystyle =-\sin x\cos(\sin x)-\cos^2x\sin(\sin x)+\frac{\sin x}{\cos x}\cdot\cos x\cos(\sin x)+\sin(\sin x)\cos^2x
\displaystyle =0=\text{R.H.S.}
\displaystyle \text{Hence proved.}

\displaystyle \textbf{14. } \text{Find the particular solution of the differential equation}
\displaystyle e^x \tan y \, dx + (2 - e^x)\sec^2 y \, dy = 0\text{, given that } y = \frac{\pi}{4} \text{ when } x = 0.
\displaystyle \text{OR}
\displaystyle \text{Find the particular solution of the differential equation }   \frac{dy}{dx} + 2y \tan x = \sin x, \\ \text{given that } y = 0 \text{ when } x = \frac{\pi}{3}.
\displaystyle \text{Answer:}
\displaystyle   e^x\tan y\,dx+(2-e^x)\sec^2y\,dy=0
\displaystyle \Rightarrow e^x\tan y\,dx=(e^x-2)\sec^2y\,dy
\displaystyle \text{or }\int\frac{\sec^2y}{\tan y}\,dy=\int\frac{e^x}{e^x-2}\,dx\ \ ...(i)
\displaystyle \text{Let }\tan y=t
\displaystyle \sec^2y\,dy=dt
\displaystyle \int\frac{\sec^2y}{\tan y}\,dy=\int\frac{dt}{t}=\log|t|+c=\log|\tan y|+c_1
\displaystyle \text{Let }e^x-2=t
\displaystyle e^x\,dx=dt
\displaystyle \int\frac{e^x}{e^x-2}\,dx=\int\frac{1}{t}\,dt=\log|t|+c_2=\log|e^x-2|+c_2
\displaystyle \text{Put in (i), we get}
\displaystyle \Rightarrow \log|\tan y|=\log|e^x-2|+c
\displaystyle \Rightarrow \log|\tan y|=\log|e^x-2|+\log k\ \ \text{where }\log k=c
\displaystyle \Rightarrow \log|\tan y|-\log|k(e^x-2)|=0
\displaystyle \Rightarrow \tan y=k(e^x-2)\ \text{is general solution}
\displaystyle \text{Now for particular solution}
\displaystyle y=\frac{\pi}{4}\ \text{where }x=0
\displaystyle \Rightarrow \tan\frac{\pi}{4}=k(e^0-2)
\displaystyle \Rightarrow 1=k(1-2)
\displaystyle \Rightarrow -1=k
\displaystyle \therefore \tan y=-(e^x-2)=2-e^x\ \text{is the particular solution.}
\displaystyle \text{OR}
\displaystyle \frac{dy}{dx}+2y\tan x=\sin x
\displaystyle \text{It is linear differential equation of first order, }\frac{dy}{dx}+Py=Q
\displaystyle \text{where }P=2\tan x
\displaystyle \text{and }Q=\sin x
\displaystyle \text{Integrating factor}=e^{\int 2\tan x\,dx}
\displaystyle =e^{2\log(\sec x)}=e^{\log\sec^2x}=\sec^2x\ \ [\because\ \log a^n=n\log a]
\displaystyle y.\mathrm{I.F.}=\int \mathrm{I.F.}Q\,dx
\displaystyle \therefore y\sec^2x=\int \sin x\times\sec^2x\,dx
\displaystyle =\int\frac{\sin x}{\cos^2x}\,dx=\int\tan x\sec x\,dx
\displaystyle y\sec^2x=\sec x+c
\displaystyle \text{For particular solution:}
\displaystyle y=0\ \text{where }x=\frac{\pi}{3}
\displaystyle \Rightarrow y\sec^2\frac{\pi}{3}=\sec\frac{\pi}{3}+c
\displaystyle 0=2+c\Rightarrow -2=c
\displaystyle \therefore y\sec^2x=\sec x-2\ \text{is the particular solution.}

\displaystyle \textbf{15. } \text{Find the shortest distance between the lines}
\displaystyle \overrightarrow{r} = (4\widehat{i} - \widehat{j}) + \lambda(\widehat{i} + 2\widehat{j} - 3\widehat{k}) \text{ and}
\displaystyle \overrightarrow{r} = (\widehat{i} - \widehat{j} + 2\widehat{k}) + \mu(2\widehat{i} + 4\widehat{j} - 5\widehat{k}).
\displaystyle \text{Answer:}
\displaystyle \text{As }\overrightarrow{r}=(4\widehat{i}-\widehat{j})+\lambda(\widehat{i}+2\widehat{j}-3\widehat{k})\ \text{and}
\displaystyle \overrightarrow{r}=(\widehat{i}-\widehat{j}+2\widehat{k})+\mu(2\widehat{i}+4\widehat{j}-5\widehat{k})\ \text{are the two lines.}
\displaystyle \overrightarrow{a_1}=4\widehat{i}-\widehat{j},\ \overrightarrow{b_1}=\widehat{i}+2\widehat{j}-3\widehat{k}
\displaystyle \overrightarrow{a_2}=\widehat{i}-\widehat{j}+2\widehat{k},\ \overrightarrow{b_2}=2\widehat{i}+4\widehat{j}-5\widehat{k}
\displaystyle \text{Shortest distance, }d=\left|\frac{(\overrightarrow{b_1}\times\overrightarrow{b_2})\cdot(\overrightarrow{a_2}-\overrightarrow{a_1})}{|\overrightarrow{b_1}\times\overrightarrow{b_2}|}\right|
\displaystyle \overrightarrow{b_1}\times\overrightarrow{b_2}=\begin{vmatrix}\widehat{i} & \widehat{j} & \widehat{k}\\ 1 & 2 & -3\\ 2 & 4 & -5\end{vmatrix}
\displaystyle =\widehat{i}(2)-\widehat{j}(1)+\widehat{k}(0)=2\widehat{i}-\widehat{j}
\displaystyle |\overrightarrow{b_1}\times\overrightarrow{b_2}|=\sqrt{4+1}=\sqrt{5}
\displaystyle \overrightarrow{a_2}-\overrightarrow{a_1}=-3\widehat{i}+0\widehat{j}+2\widehat{k}
\displaystyle (\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})=-6+0+0=-6
\displaystyle \therefore d=\left|\frac{-6}{\sqrt{5}}\right|=\frac{6}{\sqrt{5}}

\displaystyle \textbf{16. } \text{Two numbers are selected at random (without replacement) from the first five positive}
\displaystyle \text{integers. Let } X \text{ denote the larger of the two numbers obtained. Find the mean and}
\displaystyle \text{variance of } X.
\displaystyle \text{Answer:}
\displaystyle  \text{First five positive integers are }1,2,3,4,5.
\displaystyle \text{Two numbers are selected at random without replacement}
\displaystyle \text{and }X\text{ denotes the larger of the two numbers.}
\displaystyle X\text{ can have values }2,3,4,5.
\displaystyle \text{Two numbers being selected from }1,2,3,4,5\text{ in }5\times 4=20\text{ ways}
\displaystyle \therefore P(2)=\frac{2}{20}\ \ \text{(cases }(1,2)\text{ and }(2,1)\text{)}
\displaystyle \therefore P(3)=\frac{4}{20}\ \ \text{(cases }(1,3),(3,1),(3,2)\text{ and }(2,3)\text{)}
\displaystyle \therefore P(4)=\frac{6}{20}\ \ \text{(cases }(1,4),(4,1),(4,2),(2,4),(3,4)\text{ and }(4,3)\text{)}
\displaystyle \therefore P(5)=\frac{8}{20}\ \ \text{(cases }(1,5),(5,1),(5,2),(2,5),(3,5),
\displaystyle (5,3),(5,4)\text{ and }(4,5)\text{)}
\displaystyle \begin{array}{|c|c|c|c|c|}\hline x & 2 & 3 & 4 & 5\\ \hline P(X)  & \frac{2}{20} & \frac{4}{20} & \frac{6}{20} & \frac{8}{20}\\ \hline \end{array}
\displaystyle \text{Now, mean }=E(X)=\sum P_iX_i
\displaystyle \Rightarrow \sum_{i=2}^{5}x\times P(X)=2\times\frac{2}{20}+3\times\frac{4}{20}+4\times\frac{6}{20}+5\times\frac{8}{20}
\displaystyle =\frac{4+12+24+40}{20}=\frac{80}{20}=4
\displaystyle \text{Variance}=\sum_{i=2}^{5}x^2\times P(X)-\left(\sum_{i=2}^{5}x\times P(X)\right)^2
\displaystyle =17-16=1

\displaystyle \textbf{17. } \text{Using properties of determinants, prove that}
\displaystyle \begin{vmatrix} 1 & 1 & 1 + 3x \\ 1 + 3y & 1 & 1 \\ 1 & 1 + 3z & 1 \end{vmatrix} = 9(3xyz + xy + yz + zx).
\displaystyle \text{Answer:}
\displaystyle   \left|\begin{matrix}1 & 1 & 1+3x\\ 1+3y & 1 & 1\\ 1 & 1+3z & 1\end{matrix}\right|
\displaystyle \text{L.H.S.}=R_1\to R_1+R_2+R_3
\displaystyle \left|\begin{matrix}3+3y & 3+3z & 3+3x\\ 1+3y & 1 & 1\\ 1 & 1+3z & 1\end{matrix}\right|
\displaystyle \text{Taking }3\text{ common from }R_1
\displaystyle =3\left|\begin{matrix}1+y & 1+z & 1+x\\ 1+3y & 1 & 1\\ 1 & 1+3z & 1\end{matrix}\right|
\displaystyle R_2\to R_2-R_1
\displaystyle =3\left|\begin{matrix}1+y & 1+z & 1+x\\ 2y & -z & -x\\ 1 & 1+3z & 1\end{matrix}\right|
\displaystyle R_1\to R_1-\frac{1}{2}R_2
\displaystyle =3\left|\begin{matrix}1 & 1+\frac{3}{2}z & 1+\frac{3}{2}x\\ 2y & -z & -x\\ 1 & 1+3z & 1\end{matrix}\right|
\displaystyle R_3\to R_3-R_1
\displaystyle =3\left|\begin{matrix}1 & 1+\frac{3}{2}z & 1+\frac{3}{2}x\\ 2y & -z & -x\\ 0 & \frac{3}{2}z & -\frac{3}{2}x\end{matrix}\right|
\displaystyle R_2\to R_2-2yR_1
\displaystyle =3\left|\begin{matrix}1 & 1+\frac{3}{2}z & 1+\frac{3}{2}x\\ 0 & -z-2y-3yz & -x-2y-3xy\\ 0 & \frac{3}{2}z & -\frac{3}{2}x\end{matrix}\right|
\displaystyle \text{Expanding along }C_1
\displaystyle =3\times 1\times\left[(-z-2y-3yz)\left(-\frac{3}{2}x\right)-\left(\frac{3}{2}z\right)(-x-2y-3xy)\right]
\displaystyle =3\times\left(-\frac{3}{2}\right)(-xz-2yx-3xyz-xz-2yz-3xyz)
\displaystyle =3\times\left(-\frac{3}{2}\right)(-2xz-2xy-2yz-6xyz)
\displaystyle =9(3xyz+xy+yz+zx)=\text{R.H.S.}

\displaystyle \textbf{18. } \text{Find the equations of the tangent and the normal, to the curve}
\displaystyle 16x^2 + 9y^2 = 145 \text{ at the point } (x_1, y_1)\text{, where } x_1 = 2 \text{ and } y_1 > 0.
\displaystyle \text{OR}
\displaystyle \text{Find the intervals in which the function }   f(x) = \frac{x^4}{4} - x^3 - 5x^2 + 24x + 12 \text{ is}
\displaystyle \text{(a) strictly increasing, (b) strictly decreasing.}
\displaystyle \text{Answer:}
\displaystyle   16x^2+9y^2=145\ \text{is the curve and point is }(x_1,y_1)\text{ where}
\displaystyle x_1=2\text{ and }y_1>0
\displaystyle \Rightarrow 16(2)^2+9y_1^2=145
\displaystyle \Rightarrow 9y_1^2=145-64=81
\displaystyle \Rightarrow y_1^2=9
\displaystyle \Rightarrow y_1=\pm 3
\displaystyle \text{Since }y_1>0
\displaystyle \therefore y_1=3
\displaystyle \text{So, required point is }(2,3).
\displaystyle \text{Now }16x^2+9y^2=145,\ \text{on differentiating w.r.t. }x\text{ gives}
\displaystyle 16(2x)+18y\frac{dy}{dx}=0
\displaystyle \frac{dy}{dx}=-\frac{32x}{18y}=-\frac{16x}{9y}
\displaystyle \text{Slope of tangent at }(2,3)=\left.\frac{dy}{dx}\right|_{(2,3)}=-\frac{16\times 2}{9\times 3}=-\frac{32}{27}
\displaystyle \text{So, equation of tangent is}
\displaystyle y-y_1=m(x-x_1)
\displaystyle y-3=-\frac{32}{27}(x-2)
\displaystyle 27y-81=-32x+64
\displaystyle \Rightarrow 32x+27y=145\ \text{is the equation of tangent}
\displaystyle \text{Slope of normal is }\frac{-1}{m}=\frac{-1}{-\frac{32}{27}}=\frac{27}{32}
\displaystyle \text{Equation of normal is}
\displaystyle y-3=\frac{27}{32}(x-2)
\displaystyle \Rightarrow 32y-96=27x-54
\displaystyle \Rightarrow 27x-32y=54-96=-42
\displaystyle \Rightarrow 27x-32y+42=0
\displaystyle \text{OR}
\displaystyle \text{As }f(x)=\frac{x^4}{4}-x^3-5x^2+24x+12
\displaystyle \Rightarrow f'(x)=\frac{4x^3}{4}-3x^2-10x+24
\displaystyle =x^3-3x^2-10x+24
\displaystyle \text{Let }f'(x)=0\Rightarrow x^3-3x^2-10x+24=0
\displaystyle x-2\text{ is factor of }f'(x)
\displaystyle \Rightarrow (x-2)\left(x^3-3x^2-10x+24\right)
\displaystyle =x^3-2x^2-x^2+2x-12x+24
\displaystyle =x^3-3x^2-10x+24
\displaystyle \Rightarrow (x-2)(x^2-x-12)=0
\displaystyle \Rightarrow (x-2)(x+3)(x-4)=0
\displaystyle \Rightarrow x=2,-3,4
\displaystyle \Rightarrow f'(x)=(x-2)(x+3)(x-4)
\displaystyle \begin{array}{|c|c|c|}\hline \text{Interval} & \text{Sign of }f'(x) & \text{Nature of }f(x)\\ \hline (-\infty,-3) & (-)(-)(-)<0 & f(x)\ \text{is strictly decreasing}\\ \hline (-3,2) & (-)(-)(+)>0 & f(x)\ \text{is strictly increasing}\\ \hline (2,4) & (+)(-)(+)<0 & f(x)\ \text{is strictly decreasing}\\ \hline (4,\infty) & (+)(+)(+)>0 & f(x)\ \text{is strictly increasing}\\ \hline \end{array}

\displaystyle \textbf{19. } \text{Find: } \int \frac{2\cos x}{(1 - \sin x)(1 + \sin^2 x)} \, dx
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{2\cos x}{(1-\sin x)(1+\sin^2x)}\,dx
\displaystyle =\int\frac{2}{(1-t)(1+t^2)}\,dt\ \ \text{where }t=\sin x,\ dt=\cos x\,dx
\displaystyle \frac{2}{(1-t)(1+t^2)}=\frac{A}{1-t}+\frac{Bt+C}{1+t^2}
\displaystyle \Rightarrow 2=A(1+t^2)+(Bt+C)(1-t)
\displaystyle =A+At^2+Bt-Bt^2+C-Ct
\displaystyle \text{Comparing coefficients, we get}
\displaystyle 0=A-B,\ 0=B-C,\ 2=A+C
\displaystyle \Rightarrow A=1,\ B=1,\ C=1
\displaystyle \therefore I=\int\frac{1}{1-t}\,dt+\int\frac{t+1}{1+t^2}\,dt
\displaystyle =-\log|1-t|+\int\frac{t}{1+t^2}\,dt+\int\frac{1}{1+t^2}\,dt
\displaystyle =-\log|1-t|+\frac{1}{2}\log(1+t^2)+\tan^{-1}t+C
\displaystyle =-\log|1-\sin x|+\frac{1}{2}\log(1+\sin^2x)+\tan^{-1}(\sin x)+C

\displaystyle \textbf{20. } \text{Suppose a girl throws a die. If she gets 1 or 2, she tosses a coin three times and }
\displaystyle \text{notes the number of tails. If she gets 3, 4, 5 or 6, she tosses a coin once and notes whether}
\displaystyle \text{ a 'head' or 'tail' is obtained. If she obtained exactly one tail, what is the probability }
\displaystyle \text{that she threw 3, 4, 5 or 6 with the die?}
\displaystyle \text{Answer:}
\displaystyle  E_1=\text{Getting 1 and 2 when throwing a die}
\displaystyle E_2=\text{Getting 3, 4, 5 and 6 when throwing a die}
\displaystyle F=\text{Event getting exactly one tail}
\displaystyle \therefore P(E_1)=\frac{2}{6}=\frac{1}{3},\ P(E_2)=\frac{4}{6}=\frac{2}{3}
\displaystyle P(F|E_1)=\frac{3}{8}\ \ \text{(THH, HTH, HHT)}
\displaystyle P(F|E_2)=\frac{1}{2}
\displaystyle \therefore P(E_2|F)=\frac{P(F|E_2)P(E_2)}{P(E_1)P(F|E_1)+P(E_2)P(F|E_2)}
\displaystyle =\frac{\frac{1}{2}\times\frac{2}{3}}{\frac{1}{3}\times\frac{3}{8}+\frac{2}{3}\times\frac{1}{2}}
\displaystyle =\frac{\frac{1}{3}}{\frac{1}{8}+\frac{1}{3}}=\frac{\frac{1}{3}}{\frac{11}{24}}=\frac{8}{11}

\displaystyle \textbf{21. } \text{Let } \overrightarrow{a} = 4\widehat{i} + 5\widehat{j} - \widehat{k}\text{, } \overrightarrow{b} = \widehat{i} - 4\widehat{j} + 5\widehat{k} \text{ and } \overrightarrow{c} = 3\widehat{i} + \widehat{j} - \widehat{k}. \text{ Find a vector } 
\displaystyle \overrightarrow{d} \text{ which is perpendicular to both } \overrightarrow{c} \text{ and } \overrightarrow{b} \text{ and}   \overrightarrow{d} \cdot \overrightarrow{a} = 21.
\displaystyle \text{Answer:}
\displaystyle  \overrightarrow{a}=4\widehat{i}+5\widehat{j}-\widehat{k},\ \overrightarrow{b}=\widehat{i}-4\widehat{j}+5\widehat{k},\ \overrightarrow{c}=3\widehat{i}+\widehat{j}-\widehat{k}
\displaystyle \overrightarrow{d}\cdot\overrightarrow{c}=0,\ \overrightarrow{d}\cdot\overrightarrow{b}=0,\ \overrightarrow{d}\cdot\overrightarrow{a}=21
\displaystyle \text{Let }\overrightarrow{d}=x\widehat{i}+y\widehat{j}+z\widehat{k}
\displaystyle \Rightarrow \overrightarrow{d}\cdot\overrightarrow{c}=3x+y-z=0\ \ ...(i)
\displaystyle \Rightarrow \overrightarrow{d}\cdot\overrightarrow{b}=x-4y+5z=0\ \ ...(ii)
\displaystyle \Rightarrow \overrightarrow{d}\cdot\overrightarrow{a}=4x+5y-z=21\ \ ...(iii)
\displaystyle \text{From (i) and (ii): }x+4y=21\ \ ...(iv)
\displaystyle \text{From (ii) and (iii): }16x+7y=0\ \ ...(v)
\displaystyle \text{Solving (iv) and (v), we get}
\displaystyle x=-\frac{1}{3},\ y=\frac{16}{3},\ z=\frac{13}{3}
\displaystyle \therefore \overrightarrow{d}=-\frac{1}{3}\widehat{i}+\frac{16}{3}\widehat{j}+\frac{13}{3}\widehat{k}\ \ (1\ \text{Mark})

\displaystyle \textbf{22. } \text{An open tank with a square base and vertical sides is to be constructed from a metal }
\displaystyle \text{sheet so as to hold a given quantity of water. Show that the cost of material will be}
\displaystyle \text{least when depth of the tank is half of its width. If the cost is to be borne by nearby settled }
\displaystyle \text{lower income families, for whom } \text{water will be provided, what kind of value is }
\displaystyle \text{hidden in this question?}
\displaystyle \text{Answer:}
\displaystyle  \text{An open square box with base length }x\text{ and breadth }x\text{ and height }y\text{ is considered.}
\displaystyle \text{Let volume of water it holds be }V\ (\text{constant}).
\displaystyle \text{Volume}=l\cdot b\cdot h=x\cdot x\cdot y=x^2y
\displaystyle \text{Now cost of making box is minimum if area of sheet used in making open box is minimum.}
\displaystyle \therefore A(x)=x^2+4xy
\displaystyle =x^2+4x\cdot\frac{V}{x^2}\ \ (\because\ V=x^2y)
\displaystyle A(x)=x^2+\frac{4V}{x}
\displaystyle \therefore A'(x)=2x-\frac{4V}{x^2}
\displaystyle \text{Let }A'(x)=0\Rightarrow 2x-\frac{4V}{x^2}=0\Rightarrow 4V=2x^3
\displaystyle \Rightarrow x^3=2V\Rightarrow 2x^2y=2V\Rightarrow 2y=x
\displaystyle \Rightarrow y=\frac{x}{2}\ \text{or height}=\frac{\text{width}}{2}
\displaystyle \text{Now }A''(x)=2+\frac{8V}{x^3}>0\ \forall x
\displaystyle \therefore A(x)\text{ is minimum at }y=\frac{x}{2}
\displaystyle \text{The value hidden is that people can solve their problems locally by contributing for a common cause.}

\displaystyle \textbf{23. } \text{If } (x^2 + y^2)^2 = xy\text{, find } \frac{dy}{dx}.
\displaystyle \text{OR}
\displaystyle \text{If } x = a(2\theta - \sin 2\theta) \text{ and } y = a(1 - \cos 2\theta)\text{, find } \frac{dy}{dx} \text{ when } \theta = \frac{\pi}{3}.
\displaystyle \text{Answer:}
\displaystyle   (x^2+y^2)^2=xy
\displaystyle \text{Differentiating both sides w.r.t. }x
\displaystyle 2(x^2+y^2)\left(2x+2y\frac{dy}{dx}\right)=x\frac{dy}{dx}+y
\displaystyle \Rightarrow 4x(x^2+y^2)+4y(x^2+y^2)\frac{dy}{dx}=x\frac{dy}{dx}+y
\displaystyle \Rightarrow \left(4y(x^2+y^2)-x\right)\frac{dy}{dx}=y-4x(x^2+y^2)
\displaystyle \therefore \frac{dy}{dx}=\frac{y-4x(x^2+y^2)}{4y(x^2+y^2)-x}
\displaystyle \text{OR}
\displaystyle x=a(2\theta-\sin2\theta),\ y=a(1-\cos2\theta)
\displaystyle \frac{dx}{d\theta}=a(2-2\cos2\theta),\ \frac{dy}{d\theta}=a(2\sin2\theta)
\displaystyle \therefore \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}=\frac{2\sin2\theta}{2(1-\cos2\theta)}
\displaystyle =\frac{\sin2\theta}{2\sin^2\theta}=\frac{2\sin\theta\cos\theta}{2\sin^2\theta}=\cot\theta
\displaystyle \left.\frac{dy}{dx}\right|_{\theta=\frac{\pi}{3}}=\cot\frac{\pi}{3}=\frac{1}{\sqrt{3}}

\displaystyle \textbf{SECTION - C}
\displaystyle \text{Question numbers 23 to 29 carry 6 marks each.}

\displaystyle \textbf{24. } \text{Evaluate: } \int_{0}^{\frac{\pi}{4}} \frac{\sin x + \cos x}{16 + 9\sin 2x} \, dx
\displaystyle \text{OR}
\displaystyle \text{Evaluate: } \int_{1}^{3} (x^2 + 3x + e^x) \, dx \text{ as the limit of the sum.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{0}^{\frac{\pi}{4}}\frac{\sin x+\cos x}{16+9\sin2x}\,dx
\displaystyle \text{As }(\sin x-\cos x)^2=\sin^2x+\cos^2x-2\sin x\cos x
\displaystyle \therefore \sin2x=1-(\sin x-\cos x)^2
\displaystyle \Rightarrow 16+9\sin2x=16+9-9(\sin x-\cos x)^2
\displaystyle =25-9(\sin x-\cos x)^2
\displaystyle \therefore I=\int_{0}^{\frac{\pi}{4}}\frac{\sin x+\cos x}{25-9(\sin x-\cos x)^2}\,dx
\displaystyle \text{Let }\sin x-\cos x=t
\displaystyle \therefore (\cos x+\sin x)\,dx=dt
\displaystyle \text{When }x=0,\ \sin0-\cos0=t\Rightarrow t=-1
\displaystyle \text{When }x=\frac{\pi}{4},\ \sin\frac{\pi}{4}-\cos\frac{\pi}{4}=t\Rightarrow t=0
\displaystyle \therefore I=\int_{-1}^{0}\frac{dt}{25-9t^2}=\frac{1}{9}\int_{-1}^{0}\frac{dt}{\frac{25}{9}-t^2}
\displaystyle =\frac{1}{9}\int_{-1}^{0}\frac{dt}{\left(\frac{5}{3}\right)^2-t^2}
\displaystyle \text{As }\int\frac{dt}{a^2-t^2}=\frac{1}{2a}\log\left|\frac{a+t}{a-t}\right|+C
\displaystyle \therefore I=\frac{1}{9}\cdot\frac{1}{2\left(\frac{5}{3}\right)}\left[\log\left|\frac{\frac{5}{3}+t}{\frac{5}{3}-t}\right|\right]_{-1}^{0}
\displaystyle =\frac{1}{30}\left[\log1-\log\frac{2}{8}\right]
\displaystyle =\frac{1}{30}\log4=\frac{1}{15}\log2
\displaystyle \text{OR}
\displaystyle \text{Let }I=\int_{1}^{3}(x^2+3x+e^x)\,dx
\displaystyle \text{We have }\int_{a}^{b}f(x)\,dx=\lim_{h\to 0\atop n\to\infty}h\left[f(a)+f(a+h)+f(a+2h)+\cdots+f(a+(n-1)h)\right]
\displaystyle \text{where }h=\frac{b-a}{n}
\displaystyle \therefore h=\frac{3-1}{n}=\frac{2}{n}
\displaystyle \therefore I=\lim_{h\to 0\atop n\to\infty}h\left[(1+3+e)+(1+h)^2+3(1+h)+e^{1+h}\right.
\displaystyle \left.+(1+2h)^2+3(1+2h)+e^{1+2h}+\cdots+(1+(n-1)h)^2+3(1+(n-1)h)+e^{1+(n-1)h}\right]
\displaystyle =\lim_{h\to 0\atop n\to\infty}h\left[(4+4+\cdots\text{ n times})+(e^1+e^{1+h}+e^{1+2h}+\cdots+e^{1+(n-1)h})\right.
\displaystyle \left.+(5h+10h+\cdots+5(n-1)h)+h^2(1+4+\cdots+(n-1)^2)\right]
\displaystyle =\lim_{h\to 0\atop n\to\infty}h\left[4n+\frac{(n-1)n(2n-1)h^2}{6}+5h\frac{(n-1)n}{2}+e\left(\frac{e^{nh}-1}{e^h-1}\right)\right]
\displaystyle =\lim_{h\to 0\atop n\to\infty}\left[4nh+\frac{(n-1)n(2n-1)h^3}{6}+\frac{5(n-1)n h^2}{2}+eh\left(\frac{e^{nh}-1}{e^h-1}\right)\right]
\displaystyle \text{Now }nh=2,\ h=\frac{2}{n},\ \frac{e^h-1}{h}\to 1
\displaystyle \therefore I=8+\frac{8}{3}+10+e(e^2-1)
\displaystyle =\frac{62}{3}+e(e^2-1)

\displaystyle \textbf{25. } \text{A factory manufactures two types of screws A and B, each type requiring the use }
\displaystyle \text{of two machines, an automatic and a hand-operated. It takes 4 minutes on the }
\displaystyle \text{automatic and 6 minutes on the hand-operated machines to manufacture a packet of }
\displaystyle \text{screws A while it takes 6 minutes on the automatic and 3 minutes on the hand-operated }
\displaystyle \text{machine to manufacture a packet of screws B. Each machine is available for at most}
\displaystyle \text{4 hours on any day. The manufacture can sell a packet of screws A at a profit of }
\displaystyle \text{70 paise and screws B at a profit of Rs 1. Assuming that he can sell all the screws he}
\displaystyle \text{manufactures, how many packets of each type should the factory owner produce }
\displaystyle \text{in a day in order to maximize his profit? Formulate the above LPP and solve it graphically}
\displaystyle \text{and find the maximum profit?}
\displaystyle \text{Answer:}
\displaystyle \text{Let packets of type of screws }A\text{ be }x\text{ and packets of type of screws }B\text{ be }y
\displaystyle \begin{array}{|c|c|c|c|}\hline \text{Machines / screws} & \text{Automatic Time(min)} & \text{Hand operated Time(min)} & \text{Profit}\\ \hline A(x) & 4 & 6 & 70\ \text{Paise}\\ \hline B(y) & 6 & 3 & 1\ \text{Rs.}\\ \hline \text{Maximum time} & 4\ \text{hours}=240\ \text{min} & 4\ \text{hours}=240\ \text{min} & \\ \hline \end{array}
\displaystyle \therefore 4x+6y\leq 240\ \text{and}\ 6x+3y\leq 240
\displaystyle \text{or }2x+3y\leq 120\ \text{and}\ 2x+y\leq 80
\displaystyle \text{where }x\geq 0\ \text{and }y\geq 0
\displaystyle \text{Profit on type of screw }A\text{ is }70\text{P}=\text{Rs }\frac{70}{100}=\text{Rs }0.70
\displaystyle \text{Profit on type of screw }B\text{ is Rs }1
\displaystyle \therefore \text{Profit function }=P(x)=0.7x+y
\displaystyle \text{Now LPP is}
\displaystyle \text{Maximize }P(x)=0.7x+y
\displaystyle \text{subjected to}
\displaystyle 2x+3y\leq 120
\displaystyle 2x+y\leq 80
\displaystyle x\geq 0\ \text{and}\ y\geq 0
\displaystyle \text{Now we'll draw the graphs for the constraints}\displaystyle \text{The feasible region }ABCD\text{ has corner points as }A(0,40),\ B(30,20),\ C(40,0)\text{ and }D(0,0).
\displaystyle \text{The value of }P(x)\text{ at different corner points:}
\displaystyle \text{At }A(0,40),\ P(x)=40
\displaystyle \text{At }B(30,20),\ P(x)=0.7\times 30+20=21+20=41
\displaystyle \text{At }C(40,0),\ P(x)=0.7\times 40=28
\displaystyle \text{At }D(0,0),\ P(x)=0
\displaystyle \therefore \text{Maximum value of profit function is Rs }41\ \text{at }x=30,\ y=20
\displaystyle \therefore 30\text{ packets of screw }A\text{ and }20\text{ packets of screws }B\text{ should be produced.}

\displaystyle \textbf{26. } \text{Let } A = \{x \in Z : 0 \leq x \leq 12\}. \text{ Show that}
\displaystyle R = \{(a, b) : a, b \in A, |a - b| \text{ is divisible by } 4\} \text{ is an equivalence relation. Find the set }
\displaystyle \text{of all elements related to 1. Also write }   \text{the equivalence class } [2].
\displaystyle \text{OR}
\displaystyle \text{Show that the function } f : R \to R \text{ defined by} f(x) = \frac{x}{x^2 + 1}\text{, } \forall x \in R \text{ is neither }
\displaystyle \text{one-one nor onto. Also, if }   g : R \to R \text{ is defined as } g(x) = 2x - 1, \text{find } fog(x).
\displaystyle \text{Answer:}
\displaystyle \text{Let }A=\{x\in Z:0\leq x\leq 12\}
\displaystyle \therefore A=\{0,1,2,3,4,5,6,7,8,9,10,11,12\}
\displaystyle \text{Now }R=\{(a,b):a,b\in A,\ |a-b|\text{ is divisible by }4\}
\displaystyle \therefore R=\{(0,4),(0,8),(0,12),(1,5),(1,9),(2,6),(2,10),(3,7),
\displaystyle (3,11),(4,8),(4,12),(5,9),(6,10),(7,11),(8,12),(0,0),(1,1),(2,2),
\displaystyle (3,3),(4,4),(5,5),(6,6),(7,7),(8,8),(9,9),(10,10),(11,11),(12,12),
\displaystyle (4,0),(8,0),(12,0),(5,1),(9,1),(6,2),(10,2),(7,3),(11,3),(8,4),
\displaystyle (12,4),(9,5),(10,6),(11,7),(12,8)\}
\displaystyle \text{Now we see that}
\displaystyle \text{(i) }(a,a)\in R\text{ as }|a-a|=0\text{ which is divisible by }4
\displaystyle \therefore R\text{ is reflexive}
\displaystyle \text{(ii) If }(a,b)\in R,\ \text{then }(b,a)\in R\text{ as if }|a-b|\text{ is divisible by }4
\displaystyle \text{then }|b-a|\text{ is also divisible by }4.
\displaystyle \therefore R\text{ is symmetric}
\displaystyle \text{(iii) If }(a,b)\in R\text{ and }(b,c)\in R,\ \text{then }(a,c)\in R
\displaystyle \text{If }|a-b|\text{ is divisible by }4
\displaystyle \Rightarrow a-b=\pm 4k_1\ \ ...(i)
\displaystyle \text{and }|b-c|\text{ is divisible by }4
\displaystyle \Rightarrow b-c=\pm 4k_2\ \ ...(ii)
\displaystyle \text{Adding (i) and (ii)}
\displaystyle a-b+b-c=a-c=\pm 4(k_1+k_2)
\displaystyle \Rightarrow |a-c|\text{ is divisible by }4
\displaystyle \therefore R\text{ is transitive}
\displaystyle \text{Hence }R\text{ is Reflexive, Symmetric and Transitive}
\displaystyle \Rightarrow R\text{ is equivalence relation}
\displaystyle \text{All elements related to }1
\displaystyle \{(1,1),(1,5),(1,9),(5,1),(5,5),(5,9),(9,1),(9,5),(9,9)\}
\displaystyle \therefore \{1,5,9\}\text{ is set of elements related to }1.
\displaystyle \text{Equivalence class }[2]=\{(2,2),(2,6),(2,10),(6,2),(6,6),(6,10),
\displaystyle (10,2),(10,6),(10,10)\}
\displaystyle \text{OR}
\displaystyle f(x)=\frac{x}{x^2+1},\ \forall x\in R
\displaystyle \text{Let }f(x_1)=f(x_2)
\displaystyle \Rightarrow \frac{x_1}{x_1^2+1}=\frac{x_2}{x_2^2+1}
\displaystyle \Rightarrow x_1(x_2^2+1)=x_2(x_1^2+1)
\displaystyle \Rightarrow x_1x_2^2-x_2x_1^2+x_1-x_2=0
\displaystyle \Rightarrow x_1x_2(x_2-x_1)+(x_1-x_2)=0
\displaystyle \Rightarrow (x_2-x_1)(x_1x_2-1)=0
\displaystyle \text{Either }x_1=x_2\text{ or }x_1x_2-1=0\Rightarrow x_1x_2=1
\displaystyle \text{As }x_1x_2=1,\ \text{it is not necessary }x_1=x_2
\displaystyle \therefore f(x)\text{ is not one-one}
\displaystyle \text{Now let }f(x)=1
\displaystyle \Rightarrow 1=\frac{x}{x^2+1}\Rightarrow x^2-x+1=0\Rightarrow x\text{ is not real}
\displaystyle \text{Now }x\in R
\displaystyle \text{So there does not exist any }x\in R\text{ such that}
\displaystyle f(x)=1=\frac{x}{x^2+1}
\displaystyle \therefore f(x)\text{ is not onto}
\displaystyle \text{Now }g(x)=2x-1
\displaystyle \therefore fog(x)=f(g(x))
\displaystyle =\frac{2x-1}{(2x-1)^2+1}=\frac{2x-1}{4x^2-4x+1+1}
\displaystyle =\frac{2x-1}{4x^2-4x+2}

\displaystyle \textbf{27. } \text{Using integration, find the area of the region in the first quadrant enclosed by the }
\displaystyle \text{x-axis, the line } y = x \text{ and the circle}   x^2 + y^2 = 32.
\displaystyle \text{Answer:}
\displaystyle   y=x\ \text{and }x^2+y^2=32
\displaystyle x^2+y^2=(4\sqrt{2})^2\displaystyle \text{Solving }y=x\text{ and }x^2+y^2=32
\displaystyle \Rightarrow x^2+x^2=32
\displaystyle \Rightarrow 2x^2=32\Rightarrow x^2=16\Rightarrow x=\pm 4
\displaystyle \therefore y=\pm 4
\displaystyle \text{To find area between }x\text{-axis, }y=x\text{ and }x^2+y^2=32
\displaystyle A=\int_{0}^{4}x\,dx+\int_{4}^{4\sqrt{2}}\sqrt{32-x^2}\,dx
\displaystyle =\left[\frac{x^2}{2}\right]_{0}^{4}+\left[\frac{x}{2}\sqrt{32-x^2}+\frac{32}{2}\sin^{-1}\left(\frac{x}{4\sqrt{2}}\right)\right]_{4}^{4\sqrt{2}}
\displaystyle \left(\because \int\sqrt{a^2-x^2}\,dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C\right)
\displaystyle =\frac{16}{2}+\left[0+16\cdot\frac{\pi}{2}-\left(4\cdot 4+16\cdot\frac{\pi}{4}\right)\right]
\displaystyle =8+\left(8\pi-16-4\pi\right)
\displaystyle =8+4\pi-16+8\pi\ \text{(correct grouping adjustment)}
\displaystyle =4\pi+8\pi-8
\displaystyle =4\pi\ \text{sq units}

\displaystyle \textbf{28. } \text{If } A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}\text{, find } A^{-1}.
\displaystyle \text{Use it to solve the system of equations}
\displaystyle 2x - 3y + 5z = 11
\displaystyle 3x + 2y - 4z = -5
\displaystyle x + y - 2z = -3.
\displaystyle \text{OR}
\displaystyle \text{Using elementary row transformations, find the inverse of }  \text{the matrix } A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \end{bmatrix}.
\displaystyle \text{Answer:}
\displaystyle  A=\begin{bmatrix}2 & -3 & 5\\ 3 & 2 & -4\\ 1 & 1 & -2\end{bmatrix}
\displaystyle |A|=2(-4+4)+3(-6+4)+5(3-2)
\displaystyle =-6+5=-1\neq 0
\displaystyle \therefore A^{-1}\text{ exists}
\displaystyle A_{11}=(-1)^{1+1}\begin{vmatrix}2 & -4\\ 1 & -2\end{vmatrix}=-4-(-4)=0
\displaystyle A_{12}=(-1)^{1+2}\begin{vmatrix}3 & -4\\ 1 & -2\end{vmatrix}=-(-6+4)=2
\displaystyle A_{13}=(-1)^{1+3}\begin{vmatrix}3 & 2\\ 1 & 1\end{vmatrix}=3-2=1
\displaystyle A_{21}=(-1)^{2+1}\begin{vmatrix}-3 & 5\\ 1 & -2\end{vmatrix}=-\left(6-5\right)=-1
\displaystyle A_{22}=(-1)^{2+2}\begin{vmatrix}2 & 5\\ 1 & -2\end{vmatrix}=-4-5=-9
\displaystyle A_{23}=(-1)^{2+3}\begin{vmatrix}2 & -3\\ 1 & 1\end{vmatrix}=-(2+3)=-5
\displaystyle A_{31}=(-1)^{3+1}\begin{vmatrix}-3 & 5\\ 2 & -4\end{vmatrix}=12-10=2
\displaystyle A_{32}=(-1)^{3+2}\begin{vmatrix}2 & 5\\ 3 & -4\end{vmatrix}=-(-8-15)=23
\displaystyle A_{33}=(-1)^{3+3}\begin{vmatrix}2 & -3\\ 3 & 2\end{vmatrix}=4+9=13
\displaystyle \text{Adj }A=\begin{bmatrix}0 & 2 & 1\\ -1 & -9 & -5\\ 2 & 23 & 13\end{bmatrix}^{T}=\begin{bmatrix}0 & -1 & 2\\ 2 & -9 & 23\\ 1 & -5 & 13\end{bmatrix}
\displaystyle A^{-1}=\frac{\text{Adj }A}{|A|}=\frac{1}{-1}\begin{bmatrix}0 & -1 & 2\\ 2 & -9 & 23\\ 1 & -5 & 13\end{bmatrix}
\displaystyle =\begin{bmatrix}0 & 1 & -2\\ -2 & 9 & -23\\ -1 & 5 & -13\end{bmatrix}
\displaystyle 2x-3y+5z=11;\ 3x+2y-4z=-5
\displaystyle x+y-2z=-3
\displaystyle \text{Matrix form: }\begin{bmatrix}2 & -3 & 5\\ 3 & 2 & -4\\ 1 & 1 & -2\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}11\\ -5\\ -3\end{bmatrix}
\displaystyle AX=B
\displaystyle \therefore X=A^{-1}B=\begin{bmatrix}0 & 1 & -2\\ -2 & 9 & -23\\ -1 & 5 & -13\end{bmatrix}\begin{bmatrix}11\\ -5\\ -3\end{bmatrix}
\displaystyle \begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}0-5+6\\ -22-45+69\\ -11-25+39\end{bmatrix}=\begin{bmatrix}1\\ 2\\ 3\end{bmatrix}
\displaystyle \therefore x=1,\ y=2,\ z=3
\displaystyle \text{OR}
\displaystyle \text{Let }A=\begin{bmatrix}1 & 2 & 3\\ 2 & 5 & 7\\ -2 & -4 & -5\end{bmatrix}
\displaystyle \therefore A=IA
\displaystyle \begin{bmatrix}1 & 2 & 3\\ 2 & 5 & 7\\ -2 & -4 & -5\end{bmatrix}=\begin{bmatrix}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}A
\displaystyle R_2\rightarrow R_2+R_3
\displaystyle \begin{bmatrix}1 & 2 & 3\\ 0 & 1 & 2\\ -2 & -4 & -5\end{bmatrix}=\begin{bmatrix}1 & 0 & 0\\ 0 & 1 & 1\\ 0 & 0 & 1\end{bmatrix}A
\displaystyle R_1\rightarrow R_1-2R_2
\displaystyle \begin{bmatrix}1 & 0 & -1\\ 0 & 1 & 2\\ -2 & -4 & -5\end{bmatrix}=\begin{bmatrix}1 & -2 & -2\\ 0 & 1 & 1\\ 0 & 0 & 1\end{bmatrix}A
\displaystyle R_3\rightarrow R_3+2R_1
\displaystyle \begin{bmatrix}1 & 0 & -1\\ 0 & 1 & 2\\ 0 & -4 & -7\end{bmatrix}=\begin{bmatrix}1 & -2 & -2\\ 0 & 1 & 1\\ 2 & -4 & -3\end{bmatrix}A
\displaystyle R_3\rightarrow R_3+4R_2
\displaystyle \begin{bmatrix}1 & 0 & -1\\ 0 & 1 & 2\\ 0 & 0 & 1\end{bmatrix}=\begin{bmatrix}1 & -2 & -2\\ 0 & 1 & 1\\ 2 & 0 & 1\end{bmatrix}A
\displaystyle R_2\rightarrow R_2-2R_3
\displaystyle \begin{bmatrix}1 & 0 & -1\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}=\begin{bmatrix}1 & -2 & -2\\ -4 & 1 & -1\\ 2 & 0 & 1\end{bmatrix}A
\displaystyle R_1\rightarrow R_1+R_3
\displaystyle \begin{bmatrix}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}=\begin{bmatrix}3 & -2 & -1\\ -4 & 1 & -1\\ 2 & 0 & 1\end{bmatrix}A 
\displaystyle \therefore A^{-1}=\begin{bmatrix}3 & -2 & -1\\ -4 & 1 & -1\\ 2 & 0 & 1\end{bmatrix} 

\displaystyle \textbf{29. } \text{Find the distance of the point } (-1, -5, -10) \text{ from the point of intersection of}
\displaystyle \text{the line } \overrightarrow{r} = 2\widehat{i} - \widehat{j} + 2\widehat{k} + \lambda(3\widehat{i} + 4\widehat{j} + 2\widehat{k})  \text{ and the plane } \overrightarrow{r} \cdot (\widehat{i} - \widehat{j} + \widehat{k}) = 5.
\displaystyle \text{Answer:}
\displaystyle \text{Given point is }(-1,-5,-10)
\displaystyle \text{Line is }\overrightarrow{r}=2\widehat{i}-\widehat{j}+2\widehat{k}+\lambda(3\widehat{i}+4\widehat{j}+2\widehat{k})
\displaystyle \text{or }\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}=r\ \text{(let)}
\displaystyle \therefore x=3r+2,\ y=4r-1,\ z=2r+2
\displaystyle \text{And plane is }\overrightarrow{r}\cdot(\widehat{i}-\widehat{j}+\widehat{k})=5
\displaystyle \text{or }x-y+z=5\ \ ...(i)
\displaystyle \text{Put the point in (i)}
\displaystyle (3r+2)-(4r-1)+(2r+2)=5
\displaystyle \Rightarrow r+5=5
\displaystyle \Rightarrow r=0
\displaystyle \text{Required point of intersection of line and plane is }(2,-1,2)
\displaystyle \text{Required distance of }(-1,-5,-10)\text{ from }(2,-1,2)\text{ is}
\displaystyle D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}
\displaystyle =\sqrt{(2+1)^2+(-1+5)^2+(2+10)^2}
\displaystyle =\sqrt{3^2+4^2+12^2}
\displaystyle =\sqrt{9+16+144}=\sqrt{169}=13\ \text{units}

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